binary-tree

问题 I'm looking to take some code I've written in Java for a binary tree class and remove the count parameter from the arguments, but keep the whole thing recursive. So, given a class with these variables: public class BinaryTree<E> { protected E data; protected BinaryTree<E> left,right; How could I do that for: public int levelCount(int count, int level){ if (data == null) {return 0;} if (count == level) {return 1;} else { return this.getRight().levelCount(count+1,level) + this.getLeft()

问题 I'm looking to take some code I've written in Java for a binary tree class and remove the count parameter from the arguments, but keep the whole thing recursive. So, given a class with these variables: public class BinaryTree<E> { protected E data; protected BinaryTree<E> left,right; How could I do that for: public int levelCount(int count, int level){ if (data == null) {return 0;} if (count == level) {return 1;} else { return this.getRight().levelCount(count+1,level) + this.getLeft()

问题 I'm seeking ideas to plot a tuple tree t = ((4,), (3, 5,), (2, 4, 6,), (1, 3, 5, 7,)) as the following image (assuming this binomial tree size can change). I'm trying to avoid dependencies on non-core packages (just sticking to pandas, numpy, matplotlib, scikit, and such). 回答1: I'm using this piece of code which gives a pretty good result: from matplotlib import pyplot as plt import numpy as np fig = plt.figure(figsize=[5, 5]) for i in range(3): x = [1, 0, 1] for j in range(i): x.append(0) x

问题 I'm seeking ideas to plot a tuple tree t = ((4,), (3, 5,), (2, 4, 6,), (1, 3, 5, 7,)) as the following image (assuming this binomial tree size can change). I'm trying to avoid dependencies on non-core packages (just sticking to pandas, numpy, matplotlib, scikit, and such). 回答1: I'm using this piece of code which gives a pretty good result: from matplotlib import pyplot as plt import numpy as np fig = plt.figure(figsize=[5, 5]) for i in range(3): x = [1, 0, 1] for j in range(i): x.append(0) x

问题 The website GeeksforGeeks has presented a solution for the problem of Maximum path sum for a binary tree. The question is as follows: Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. The core of the solution is as follows: int findMaxUtil(Node node, Res res) { if (node == null) return 0; // l and r store maximum path sum going through left and // right child of root respectively int l = findMaxUtil(node.left, res); int r = findMaxUtil(node

问题 The website GeeksforGeeks has presented a solution for the problem of Maximum path sum for a binary tree. The question is as follows: Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. The core of the solution is as follows: int findMaxUtil(Node node, Res res) { if (node == null) return 0; // l and r store maximum path sum going through left and // right child of root respectively int l = findMaxUtil(node.left, res); int r = findMaxUtil(node

来源： https://stackoverflow.com/questions/36802354/print-binary-tree-in-a-pretty-way-using-c

来源： https://stackoverflow.com/questions/36802354/print-binary-tree-in-a-pretty-way-using-c

来源： https://stackoverflow.com/questions/14953979/find-the-diameter-of-a-binary-tree

来源： https://stackoverflow.com/questions/61681299/time-complexity-of-cloning-a-binary-tree